Complete Calculus Revision for JEE Mains and Advanced
Welcome to this detailed calculus revision guide! This post covers all key topics required for JEE Mains and Advanced, from Relations and Functions to Differential Equations. Each section includes formulas, examples, and illustrations to ensure conceptual clarity.
1. Relations and Functions
This chapter forms the basis of calculus. Key topics include:
| Topic | Explanation |
|---|---|
| Domain and Range | For a function f(x), the domain is the set of all valid inputs, and the range is the set of outputs. |
| Inverse Functions | If f(x) is bijective, then its inverse f^{-1}(x) satisfies f(f^{-1}(x)) = x. |
Illustration: Domain, Range, and Inverse Functions
Example:
For f(x) = x^2, the domain is all real numbers (-\infty, \infty), and the range is [0, \infty) because the output is always non-negative.
2. Applications of Derivatives
Derivatives are used to solve problems related to rates of change, tangents, and optimization.
- Rate of Change: The derivative represents the rate of change of a quantity: \[ \text{Rate} = \frac{dy}{dx}. \]
- Tangents and Normals: For a curve y = f(x): \[ \text{Slope of tangent} = \frac{dy}{dx}, \quad \text{Slope of normal} = -\frac{1}{\frac{dy}{dx}}. \]
- Maxima and Minima: Critical points are found where: \[ \frac{dy}{dx} = 0, \quad \text{and use } \frac{d^2y}{dx^2} \text{ to classify.} \]
Example:
For the function f(x) = x^2 - 4x + 3, find the slope of the tangent at x = 2. The derivative is f'(x) = 2x - 4. At x = 2, the slope is f'(2) = 0, indicating a minimum point.
Illustration: Tangent and Normal to a Curve
3. Limits, Continuity, and Differentiability
Understanding these concepts is critical for solving calculus problems:
- Limit: The value that f(x) approaches as x approaches a: \[ \lim_{x \to a} f(x) = L. \]
- Continuity: A function f(x) is continuous at x = c if: \[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c). \]
- Differentiability: A function is differentiable if its derivative exists at all points: \[ \frac{dy}{dx} \text{ is defined}. \]
Example:
Consider the function f(x) = \frac{1}{x}. The limit as x \to 0 does not exist, which implies that f(x) is not continuous at x = 0, hence not differentiable there.
Illustration: Continuous and Differentiable Functions
4. Differentiation
The derivative measures the instantaneous rate of change. Important formulas include:
- \frac{d}{dx}[x^n] = nx^{n-1}
- \frac{d}{dx}[\sin x] = \cos x, \quad \frac{d}{dx}[\cos x] = -\sin x
- \frac{d}{dx}[\ln x] = \frac{1}{x}
Example:
For the function f(x) = x^3 - 5x + 2, the derivative is f'(x) = 3x^2 - 5. The slope of the tangent at x = 1 is f'(1) = 3(1)^2 - 5 = -2.
5. Indefinite Integration
Integration is the reverse of differentiation:
- \int x^n dx = \frac{x^{n+1}}{n+1} + C
- \int e^x dx = e^x + C
- \int \sin x dx = -\cos x + C
Example:
To integrate \int (2x + 3) dx, the result is x^2 + 3x + C, where C is the constant of integration.
6. Definite Integration
Calculate the area under a curve using: \[ \int_a^b f(x) dx = F(b) - F(a), \] where F(x) is the antiderivative of f(x).
Example:
For f(x) = x^2, the definite integral from 0 to 2 is: \[ \int_0^2 x^2 dx = \left[ \frac{x^3}{3} \right]_0^2 = \ \[ \int_0^2 x^2 dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3}. \] This represents the area under the curve of y = x^2 from x = 0 to x = 2.
7. Application of Integrals (Area under Curve)
Integration is commonly used to find areas under curves. The formula is: \[ \text{Area} = \int_a^b f(x) \, dx. \] For example, to find the area under the curve of y = x^2 between x = 0 and x = 2, we already found that the integral is \frac{8}{3}.
Example:
Consider the function f(x) = \sqrt{x}. The area under the curve from x = 1 to x = 4 is calculated as: \[ \int_1^4 \sqrt{x} \, dx = \left[ \frac{2x^{3/2}}{3} \right]_1^4 = \frac{2(8)}{3} - \frac{2(1)}{3} = \frac{14}{3}. \]
Illustration: Area under the curve of y = \sqrt{x} between x = 1 and x = 4.
8. Differential Equation
A differential equation is an equation that involves derivatives. These are used to model various phenomena such as growth, decay, and motion.
- The general form of a first-order differential equation is: \[ \frac{dy}{dx} = f(x, y). \]
- A simple solution approach involves separation of variables, where we can separate the variables x and y and then integrate both sides: \[ \frac{dy}{dx} = g(x) \cdot h(y) \quad \Rightarrow \quad \frac{1}{h(y)} \, dy = g(x) \, dx. \]
Example:
Consider the differential equation: \[ \frac{dy}{dx} = x \cdot y. \] We separate the variables: \[ \frac{dy}{y} = x \, dx. \] Integrating both sides: \[ \int \frac{1}{y} \, dy = \int x \, dx \quad \Rightarrow \quad \ln |y| = \frac{x^2}{2} + C. \] Thus, the solution is: \[ y = \pm e^{\frac{x^2}{2} + C}. \]
Illustration: General Solution of the Differential Equation \frac{dy}{dx} = x \cdot y.
Conclusion
By understanding and practicing the key concepts of calculus, you will be well-prepared for JEE Mains and Advanced. Keep revising the formulas and solving example problems to strengthen your grasp on the subject. Make use of visual aids and simulations to improve your problem-solving skills.
Stay consistent and practice regularly to ace the calculus section in your exams!
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